Re: [PATCH] bpf: task_group_seq_get_next: cleanup the usage of next_thread()

Kui-Feng Lee
2023-08-21T20:24:08Z

On 8/21/23 11:34, Oleg Nesterov wrote:
> On 08/21, Kui-Feng Lee wrote:
>>
>>
>> On 8/21/23 08:09, Oleg Nesterov wrote:
>>> 1. find_pid_ns() + get_pid_task() under rcu_read_lock() guarantees that we
>>> can safely iterate the task->thread_group list. Even if this task exits
>>> right after get_pid_task() (or goto retry) and pid_alive() returns 0 >
>>> Kill the unnecessary pid_alive() check.
>>
>> This function will return next_task holding a refcount, and release the
>> refcount until the next time calling the same function. Meanwhile,
>> the returned task A may be killed, and its next task B may be
>> killed after A as well, before calling this function again.
>> However, even task B is destroyed (free), A's next is still pointing to
>> task B. When this function is called again for the same iterator,
>> it doesn't promise that B is still there.
>
> Not sure I understand...
>
> OK, if we have a task pointer with incremented refcount and do not hold
> rcu lock, then yes, you can't remove the pid_alive() check in this code:
>
> rcu_read_lock();
> if (pid_alive(task))
> do_something(next_thread(task));
> rcu_read_unlock();
>
> because task and then task->next can exit and do call_rcu(delayed_put_task_struct)
> before we take rcu_read_lock().
>
> But if you do something like
>
> rcu_read_lock();
>
> task = find_task_in_some_rcu_protected_list();
> do_something(next_thread(task));
>
> rcu_read_unlock();
>
> then next_thread(task) should be safe without pid_alive().
>
> And iiuc task_group_seq_get_next() always does
>
> rcu_read_lock(); // the caller does lock/unlock
>
> task = get_pid_task(pid, PIDTYPE_PID);
> if (!task)
> return;
>
> next_task = next_thread(task);
>
> rcu_read_unlock();
>
> Yes, both task and task->next can exit right after get_pid_task(), but since
> can only happen after we took rcu_read_lock(), delayed_put_task_struct() can't
> be called until we drop rcu lock.
>
> What have I missed?

Then, it makes sense to me! Thank you for the explanation.

>
> Oleg.
>